We answer in the **negative** the question as to whether there is an $R$-module isomorphism $$I \oplus J \simeq R \oplus IJ$$ whenever $I$ and $J$ are **non-invertible** ideals of a non-maximal order $R$ of some quadratic **number field** $K$.

Let
$$
F \xrightarrow[]{\varphi} G \rightarrow M \rightarrow 0
$$
be an exact sequence such that $F$ and $G$ are free modules over $R$ and $G$ is finitely generated. Let $I_i(\varphi)$ the ideal of
$R$ generated by the $i \times i$ minors of the matrix of $\varphi: F \rightarrow G$, agreeing that $I_i(\varphi) = R$ if $i \le 0$.

Then the $(i + 1)$-th *Fitting ideal* of $M$ ($i \ge 0$) is defined as $$\text{Fitt}_i(M) := I_{\text{rank}(G) - i}(\varphi).$$

The Fitting ideals are independent of $\varphi$ by Fitting's lemma [1, Corollary 20.4], i.e., they are invariant under isomorphisms of $R$-modules.

**Proposition.** Let $I, J$ be two non-zero ideals of an integral domain $R$. Assume moreover that each of $I, J$ and $IJ$ is $2$-generated. Then we have

- $\text{Fitt}_2(I \oplus J) = \text{Fitt}_1(I) \text{Fitt}_1(J)$ and $\text{Fitt}_3(I \oplus J) = \text{Fitt}_1(I) + \text{Fitt}_1(J)$

while

- $\text{Fitt}_2(R \oplus IJ) = \text{Fitt}_1(IJ)$ and $\text{Fitt}_3(R \oplus IJ) = R$.

*Proof.* Observe that $\text{Fitt}_0(I) = \text{Fitt}_0(J) = \{0\}$ [1, Proposition 20.7.a] and $I_r(\varphi \oplus \psi) = \sum_{s + t = r} I_s(\varphi) I_t(\psi)$ (use for instance [1, Definition 20.2]).

**Corollary.** Let $I$ be a $2$-generated non-zero ideal of an integral domain $R$.
Assume moreover that $I$ is not invertible and that $I^2 := I \cdot I$ is $2$-generated. Then
$I \oplus I \not\simeq R \oplus I^2$.

*Poof.* Because $I$ is not invertible, it follows from [1, Proposition 20.8], that $\text{Fitt}_1(I) \neq R$. By the above Proposition, we infer that
$\text{Fitt}_3(I \oplus I) = \text{Fitt}_1(I) \neq R = \text{Fitt}_3(R \oplus I^2)$.

For a concrete instance, we can set $$R = \mathbb{Z} + 2i \mathbb{Z}, \, I = 2\mathbb{Z}[i].$$
It is easily checked that $I^{-1} = \mathbb{Z}[i]$ so that $I$ is not invertible and $\text{Fitt}_1(I) = II^{-1} = I$. As $I \simeq I^2$,
we have in addition $\text{Fitt}_2(R \oplus I^2) = \text{Fitt}_1( I^2) = \text{Fitt}_1(I) = I$ whereas $\text{Fitt}_2(I \oplus I) = I^2 \neq I$ because of the above Proposition.

We conclude with a general positive result relating direct sum of ideals to products.

**Theorem [2, C1].** Let $R$ be an integral domain and let $I_1, \dots, I_r$ be ideals of $R$. Set $M = I_1 \oplus \cdots \oplus I_r$ and let $T$ be the torsion submodule of $\Lambda^r(M)$. Then $\Lambda^r(M) /T \simeq I_1 \cdots I_r$. Therefore if $J_1, \dots, J_r$ are ideals of $R$ such that $I_1 \oplus \cdots \oplus I_r \simeq J_1 \oplus \cdots \oplus J_r$, then $I_1 \cdots I_r \simeq J_1 \cdots J_r$.

- [1] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 2004.
- [2] H. Matsumura, "Commutative Ring Theory", 1989.